class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        # 初始化DP表（(m+1) x (n+1)）
        dp = [[False] * (n + 1) for _ in range(m + 1)]

        # 空字符串和空模式匹配
        dp[0][0] = True

        # 处理空字符串与带 '*' 的模式匹配
        for j in range(2, n + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 2]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j - 1] == s[i - 1] or p[j - 1] == '.':
                    # 当前字符匹配，状态继承
                    dp[i][j] = dp[i - 1][j - 1]
                elif p[j - 1] == '*':
                    # '*' 可能代表0个或多个
                    # 零个前字符
                    dp[i][j] = dp[i][j - 2]
                    # 多个前字符（如果匹配的话）
                    if p[j - 2] == s[i - 1] or p[j - 2] == '.':
                        dp[i][j] |= dp[i - 1][j]

        return dp[m][n]
